Solve the quadratic equation by completing the square

[chijioke]

Solve the quadratic equation by completing.
[math]y ^ { 2 } - 6 y + 9 = 0[/math]

Solution​

[math]y ^ { 2 } - 6 y = -9[/math][math]y ^ { 2 } - 6 y +\left(\frac{1}{2}×(-6)\right) ^2= -9+\left(\frac{1}{2}×(-6)\right) ^2[/math][math]y ^ { 2 } - 6+9 y = -9+9[/math][math]y ^ { 2 } - 6 y +9 = 0[/math][math]\left(y-3\right)^2=0[/math][math]y-3 = \pm\sqrt{0}[/math][math]y-3=0[/math][math]\therefore ~ y=3[/math]Is y really 3?
I am asking this question because if you solve this problem by factorization method, you get y = 3 twice.
My concern here is why am I getting y = 3 only once instead of twice.

 

[Steven G]

(y-3)^2 = (y-3)(y-3)=0. So yes, you are getting y= 3 twice.
The 6th line from the bottom is wrong.

 

[Dr.Peterson]

I am asking this question because if you solve this problem by factorization method, you get y = 3 twice.
My concern here is why am I getting y = 3 only once instead of twice.

You did in fact get y = 3 twice this way: y = 3 + 0 and y = 3 - 0. If the discriminant were not 0, these would have been two different roots; here, they are equal.

Completing the square doesn't reveal the double (multiplicity 2) nature of the root in the same way as factoring, but it is there.

 

[NotJeffM]

I am asking this question because if you solve this problem by factorization method, you get y = 3 twice.
My concern here is why am I getting y = 3 only once instead of twice.

Good question.

Please consider the following

[math]a^2 = b \iff a * a = b.[/math]
That is a definition.

[math]+ 0 \equiv 0 \equiv - 0 \implies \pm 0 = 0.[/math]
That proposition is true for no real number other than zero.

Put those two important trivialities together with your [imath](y - 3)^2 = \pm 0[/imath] to get

[math] (y - 3)^2 = \pm 0 \implies \\ (y - 3)^2 = 0 \implies \\ (y - 3)(y - 3) = 0. [/math]
Exactly the same conclusion derived from factoring.

There is something called the fundamental theorem of algebra, which says

[math] ax^2 + bx + c = 0 \text { and } a, \ b, \ c \text { are all real numbers} \implies \\ \exists \text { numbers } p \text { and } q \text { such that }\\ a(x - p)(x - q) = 0 \implies x = p \text { or } x = q. [/math]
Notice what the theorem does NOT say. It does not say that p and q are real numbers; they may be complex, conjugate numbers. Nor does it say that p and q are different numbers; they may be the same number. If

[math]p = q \implies a(x - p)(x - q) = 0 \implies a(x - p)^2 = 0 \implies x =p.[/math]
Putting that theorem into English gives this:

A quadratic in x with real coefficients can always be factored into two linear factors in x, but those two factors may be duplicates.

 

[chijioke]

(y-3)^2 = (y-3)(y-3)=0. So yes, you are getting y= 3 twice.
The 6th line from the bottom is wrong.

You said the sixth line from the bottom is wrong. How is it wrong? This is the sixth line from the bottom: [math]y ^ { 2 } - 6 y +\left(\frac{1}{2}×(-6)\right) ^2= -9+\left(\frac{1}{2}×(-6)\right) ^2[/math] . How is it wrong?
Maybe I can explain. The first line which happens to be the fifth line from the bottom gave rise to the sixth line.
The original equation is
[math]y ^ { 2 } - 6 y + 9 = 0[/math]To get the fifth line, I subtracted 9 from both sides of the equations and here comes the fifth line which is
[math]y ^ { 2 } - 6 y = -9[/math]To get the sixth line, I added the square of of half of the coefficient of y. The coefficient of y is -6. Half of it is [math]\frac{1}{2}×(-6)=-3[/math] Square of -3 is [math](-3)^2=9[/math]Here comes the sixt line
[math]y ^ { 2 } - 6 y +\left(\frac{1}{2}×(-6)\right) ^2= -9+\left(\frac{1}{2}×(-6)\right) ^2[/math]I think what is wrong is the seventh line which is supposed to be [math]y ^ { 2 } - 6y+9 = -9+9[/math]which I mistakenly wrote as
[math]y ^ { 2 } - 6+9 y = -9+9[/math]

 

[Steven G]

You said the sixth line from the bottom is wrong. How is it wrong? This is the sixth line from the bottom: [math]y ^ { 2 } - 6 y +\left(\frac{1}{2}×(-6)\right) ^2= -9+\left(\frac{1}{2}×(-6)\right) ^2[/math] . How is it wrong?
Maybe I can explain. The first line which happens to be the fifth line from the bottom gave rise to the sixth line.
The original equation is
[math]y ^ { 2 } - 6 y + 9 = 0[/math]To get the fifth line, I subtracted 9 from both sides of the equations and here comes the fifth line which is
[math]y ^ { 2 } - 6 y = -9[/math]To get the sixth line, I added the square of of half of the coefficient of y. The coefficient of y is -6. Half of it is [math]\frac{1}{2}×(-6)=-3[/math] Square of -3 is [math](-3)^2=9[/math]Here comes the sixt line
[math]y ^ { 2 } - 6 y +\left(\frac{1}{2}×(-6)\right) ^2= -9+\left(\frac{1}{2}×(-6)\right) ^2[/math]I think what is wrong is the seventh line which is supposed to be [math]y ^ { 2 } - 6y+9 = -9+9[/math]which I mistakenly wrote as
[math]y ^ { 2 } - 6+9 y = -9+9[/math]

What you are stating is not the line I am referring to. It is the line above the one you referred to.

 

[chijioke]

What you are stating is not the line I am referring to. It is the line above the one you referred to.

I myself don't even know the line you are talking about. I think it would be helpful if you show the line.

 

[Steven G]

y^2−6+9y=−9+9

 

[chijioke]

y^2−6+9y=−9+9

Yes. Thank you but I have already acknowledged that in post # 5. Didn't you see it?

I think what is wrong is the seventh line which is supposed to be [math]y ^ { 2 } - 6y+9 = -9+9[/math]which I mistakenly wrote as
[math]y ^ { 2 } - 6+9 y = -9+9[/math]